I came across the following equivalence. Let $g_j$ for $j=1,\ldots,k$ be closed convex functions. For $\lambda_j > 0$ for $j=1,\ldots,k$. Then, the term $$ \big( \sum_{j=1}^k \lambda_j g_j \big)^*(z) $$ is equivalent to

$$ \inf_{\sum z_j = z} \left\{ \sum_{j=1}^k \lambda_j g_j^* \left( \frac{z_j}{\lambda_j} \right) \right\}. $$

It is noted that this is due to the infimal convolution definition applied on the conjugate of the sum of convex functions. I don't see how though. The classical definition as in this Math.SE question is not applicable to this concept, or at least I was not able to show.

How can I show this equivalence, and is there any source online (specifically for this)?


1 Answer 1


Ok, I solved. here they show that $(\sum\lambda_j g_j)^* =(\lambda_1g_1)^* \Delta (\lambda_2g_2)^*\Delta \ldots $ where $\Delta$ is the infimal convolution operator.

Next, apply the definition of inf conv to have $\inf \{ \sum (\lambda_j g_j)^* (z_j) \}$ or equivalently $\inf \{ \sum \lambda_j (g_j)^* (z_j/\lambda_j) \}$ where $\sum z_j= z$.

  • 1
    $\begingroup$ the book on Convex Analysis by Rockafellar is a good reference for these identities $\endgroup$
    – LinAlg
    Oct 15, 2019 at 1:21

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